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Four equal charges of value +Q are place...

Four equal charges of value +Q are placed at any four vertices of a regular hexagon of side a. By suitably choosing the vertices, what can be the maximum possible magnitude of electric field at the centre of the hexagon?

(A) `(Q)/(4piin_(0)a^(2))`
(B) `(sqrt2Q)/(4piin_(0)a^(2))`
(C) `(sqrt3Q)/(4piin_(0)a^(2))`
(D) `(2Q)/(4piin_(0)a^(2))`

A

`(Q)/(4piin_(0)a^(2))`

B

`(sqrt2Q)/(4piin_(0)a^(2))`

C

`(sqrt3Q)/(4piin_(0)a^(2))`

D

`(2Q)/(4piin_(0)a^(2))`

Text Solution

Verified by Experts

If the four +Q charges are placed at four adjacent vertices, then the net electric field at point O will be maximum and the electric field `(E_(1))` due to each charge will be equal in magnitude.

Since, ABCDEF, is a regular hexagon,
So, `AO=BO=CO=DO=a`
The net electric field at point O due to +Q charges placed at points A and D is zero. Hence the net electric field at point O due to +Q charges placed at points B and C is the maximum values of the electric field.
Now, `E_(1)=(Q)/(4piin_(0)a^(2))`
`:.E'=sqrt(E_(1)^(2)+E_(1)^(2)+2E_(1)E_(1)cos60^(@))=(sqrt3Q)/(4piin_(0)a^(2))`
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